（鄂爾多斯專版）2020年中考數(shù)學(xué)復(fù)習(xí) 提分專練01 數(shù)與式的運(yùn)算

提分專練(一)數(shù)與式的運(yùn)算|類型1|實數(shù)的混合運(yùn)算1.計算:-13-2-(-1)2019-5+(-1)0.2.計算:27+1-3+12-1-20190.3.計算:(-10)0+|2-1|+12-1-2sin45°.4.計算:-1+8·cos45°-12-2+(-3.14)0.5.計算:(3-)0+4sin45°-8+1-3.6.計算:|-3|+3·tan30°-38-(2020-)0+12-1.7.計算:4-(-2020)0+|3-2|+2sin60°.8.計算:20200+2|1-sin30°|-13-1+16.|類型2|分式的化簡求值9.2019·煙臺 先化簡x+3-7x-3÷2x2-8xx-3,再從0x4中選一個適合的整數(shù)代入求值.10.先化簡,后求值:1+1x÷x2+2x+1x,其中x滿足x2-x-2=0.11.先化簡,再求值:1x-y+2x2-xy÷x+22x,其中實數(shù)x,y滿足y=x-2-4-2x+1.12.先化簡,再求值:1-2x-1·x2-xx2-6x+9,其中x是從1,2,3中選取的一個合適的數(shù).13.先化簡,再求值:x2-2x+1x2+x÷1-2x+1,其中x=3.14.先化簡,再求值:a+1-4a-5a-1÷1a-1a2-a,其中a=12-1+tan60°.15.先化簡,再求值:x2-yx-x-1÷x2-y2x2-2xy+y2,其中x=2,y=6.【參考答案】1.解:原式=9+1-5+1=6.2.解:原式=33+3-1+2-1=43.3.解:原式=1+2-1+2-2×22=2.4.解:原式=-1+22×22-4+1=-1+2-4+1=-2.5.解:原式=1+4×22-22+3-1=3.6.解:原式=3+3×33-2-1+2=3+1-2-1+2=3.7.解:原式=2-1+2-3+2×32=2-1+2-3+3=3.8.解:原式=1+2×1-12-3+4=1+2×12+1=1+1+1=3.9.解:x+3-7x-3÷2x2-8xx-3=(x+3)(x-3)x-3-7x-3×x-32x2-8x=(x+4)(x-4)x-3×x-32x(x-4)=x+42x.因為x-30,2x2-8x0,2x0,所以x不能取0,3,4,考慮到從0x4中選一個整數(shù),故x只能取1或2,當(dāng)x=1時,原式=1+42×1=52;當(dāng)x=2時,原式=2+42×2=32.(注意:與只寫一種即可)10.解:原式=x+1x·x(x+1)2=1x+1,解方程x2-x-2=0,得x1=-1,x2=2,當(dāng)x=-1時,原分式無意義.當(dāng)x=2時,原式=12+1=13.11.解:原式=x+2x(x-y)·2xx+2=2x-y,y=x-2-2(2-x)+1,x-20,2-x0,即x-2=0.x=2,y=1.故原式=22-1=2.12.解:原式=x-3x-1·x(x-1)(x-3)2=xx-3.當(dāng)x=2時,原式=22-3=-2.13.解:原式=(x-1)2x(x+1)÷x-1x+1=(x-1)2x(x+1)·x+1x-1=x-1x,當(dāng)x=3時,原式=3-13=3-33.14.解:原式=a2-1-4a+5a-1÷a-1-1aa-1=a2-4a+4a-1·aa-1a-2=a(a-2),a=12-1+tan60°=2+3,把a(bǔ)=2+3代入上式,得原式=(2+3)×3=3+23.15.解:原式=x2-yx-x2x-xx·(x-y)2(x+y)(x-y)=-y-xx·x-yx+y=-x-yx,把x=2,y=6代入得,原式=-2-62=-1+3.8